Questions: Use the standard normal distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a recent season, the population standard deviation of the yards per carry for all running backs was 1.29. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed. Which distribution should be used to construct the confidence interval? A. Use a normal distribution because n<30, the data are normally distributed and σ is unknown. B. Use a normal distribution because σ is known and the data are normally distributed. C. Use a t-distribution because n<30 and σ is unknown. D. Use a t-distribution because n<30 and σ is known. E. Cannot use the standard normal distribution or the t-distribution because σ is unknown, n<30, and the data are not normally distributed.

Use the standard normal distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results.
In a recent season, the population standard deviation of the yards per carry for all running backs was 1.29. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed.

Which distribution should be used to construct the confidence interval?
A. Use a normal distribution because n<30, the data are normally distributed and σ is unknown.
B. Use a normal distribution because σ is known and the data are normally distributed.
C. Use a t-distribution because n<30 and σ is unknown.
D. Use a t-distribution because n<30 and σ is known.
E. Cannot use the standard normal distribution or the t-distribution because σ is unknown, n<30, and the data are not normally distributed.
Transcript text: Use the standard normal distribution or the $t$-distribution to construct a $99 \%$ confidence interval for the population mean. Justify your decision. If neither distribution c be used, explain why. Interpret the results. In a recent season, the population standard deviation of the yards per carry for all running backs was 1.29 . The yards per carry of 25 randomly selected running bacl are shown below. Assume the yards per carry are normally distributed. Which distribution should be used to construct the confidence interval? A. Use a normal distribution because $n<30$, the data are normally distributed and $\sigma$ is unknown. B. Use a normal distribution because $\sigma$ is known and the data are normally distributed. C. Use a t-distribution because $\mathrm{n}<30$ and $\sigma$ is unknown. D. Use a t-distribution because $\mathrm{n}<30$ and $\sigma$ is known. E. Cannot use the standard normal distribution or the $t$-distribution because $\sigma$ is unknown, $\mathrm{n}<30$, and the data are not normally distributed.
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Solution

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Solution Steps

Step 1: Calculate the Sample Mean

The sample mean (\(\bar{x}\)) of the yards per carry for the selected running backs is calculated as follows:

\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{113.7}{24} \approx 4.7375 \]

Step 2: Determine the Distribution to Use

Since the population standard deviation (\(\sigma\)) is known and the data are normally distributed, we will use the normal distribution to construct the confidence interval. Thus, the choice is:

\[ \text{Distribution Choice: B. Use a normal distribution because } \sigma \text{ is known and the data are normally distributed.} \]

Step 3: Calculate the Confidence Interval

To calculate the 99% confidence interval, we use the formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

  • \(\bar{x} \approx 4.7375\)
  • \(z\) for 99% confidence level is approximately \(2.58\)
  • \(\sigma = 1.29\)
  • \(n = 24\)

Substituting the values:

\[ 4.7375 \pm 2.58 \cdot \frac{1.29}{\sqrt{24}} \]

Calculating the margin of error:

\[ \frac{1.29}{\sqrt{24}} \approx 0.2630 \]

Thus, the margin of error is:

\[ 2.58 \cdot 0.2630 \approx 0.6785 \]

Now, we can find the confidence interval:

\[ (4.7375 - 0.6785, 4.7375 + 0.6785) \approx (4.0590, 5.4160) \]

Final Answer

The 99% confidence interval for the population mean is:

\[ \boxed{(4.06, 5.42)} \]

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